Three Dimensional Geometry

1Direction cosines and ratios

A line's direction cosines l,m,n satisfy l^2+m^2+n^2=1; direction ratios are proportional to them. The angle between two lines follows from \cos\theta=l_1l_2+m_1m_2+n_1n_2, or the same formula with direction ratios in place of cosines once you normalise by the two magnitudes.

Two directed lines in space with direction ratios marked; \cos\theta is their normalised dot product — the spatial angle itself is what cannot be drawn honestly.

Angle from direction ratios

1. Form the dot productCompute a_1a_2+b_1b_2+c_1c_2.
2. Form the two magnitudes|\vec{d_1}|=\sqrt{a_1^2+b_1^2+c_1^2} and likewise for \vec{d_2}.
3. Divide\cos\theta is the dot product over the product of magnitudes — no need to convert to cosines first.

2Equation of a line in space

A line through \vec{a} with direction \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b}, or in Cartesian form \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}. Two lines are parallel when their direction ratios are proportional, and perpendicular when the dot product of those ratios is zero — no need to normalise first.

A point (x_1,y_1,z_1) and a direction (a,b,c) determine the line; imagine the parameter \lambda sliding along that direction in space.

Writing a line

1. Pick a pointAny known point on the line becomes (x_1,y_1,z_1) or \vec{a}.
2. Pick a directionDirection ratios (a,b,c) or vector \vec{b} — differences of two points work.
3. Equate the parametersSet \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c} (or the vector form).

3Equation of a plane

A plane with normal \vec{n} through \vec{a} is \vec{r}\cdot\vec{n}=\vec{a}\cdot\vec{n}, or ax+by+cz=d. The normal direction is (a,b,c); the intercept form is \dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=1. A line lies in the plane when a point of the line satisfies the plane and the direction is perpendicular to the normal.

Figure. Not a general plane in space: only the intercept triangle in the first octant for \dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=1, with the three axis intercepts labelled. Depth is schematic.

Reading a plane

1. Normal from the coefficientsIn ax+by+cz+d=0, the normal is (a,b,c).
2. Intercept formDivide through so the constant is 1; the intercepts are the denominators p,q,r.
3. Line in planeNeed both membership of one point and \vec{b}\perp\vec{n}.

4Angle between planes

The angle between two planes equals the angle between their normals: \cos\theta=\dfrac{\vec{n_1}\cdot\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}. Parallel planes have proportional normals; perpendicular planes have \vec{n_1}\cdot\vec{n_2}=0.

Via the normals

1. Read the normalsFrom each plane equation, take \vec{n_1} and \vec{n_2} as the coefficient triples (a,b,c).
2. Cosine of the angle\cos\theta=\dfrac{\vec{n_1}\cdot\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}. Parallel planes give proportional normals; \vec{n_1}\cdot\vec{n_2}=0 means the planes are perpendicular.
3. Report the acute one when askedIf the cosine comes out negative, the obtuse dihedral is the supplement — many stems want the acute angle between the planes.

5Distance from a point to a plane

The distance of (x_1,y_1,z_1) from ax+by+cz+d=0 is \dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}. The denominator is the magnitude of the normal — the same length that appears in every plane-angle formula.

Point-to-plane distance

1. Plug into SEvaluate |ax_1+by_1+cz_1+d|.
2. Divide by |\vec{n}|Divide by \sqrt{a^2+b^2+c^2}.
3. Keep the absolute valueDistance is positive on either side of the plane.

6Angle between a line and a plane

If the line has direction \vec{b} and the plane has normal \vec{n}, the angle \phi between line and plane satisfies \sin\phi=\dfrac{|\vec{b}\cdot\vec{n}|}{|\vec{b}||\vec{n}|}. That angle is the complement of the angle between the line and the normal — so a line parallel to the plane has \vec{b}\cdot\vec{n}=0 and \phi=0.

A line meeting a plane; \phi is the elevation from the plane to the line, the complement of the angle the line makes with the normal.

7Shortest distance between skew lines

Skew lines are neither parallel nor intersecting. The shortest distance between \vec{r}=\vec{a_1}+\lambda\vec{b_1} and \vec{r}=\vec{a_2}+\mu\vec{b_2} is \dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} — the absolute scalar projection of the join of the two points onto the common perpendicular.

Skew distance

1. Confirm skewThe lines are not parallel (\vec{b_1}\times\vec{b_2}\neq\vec{0}) and do not intersect — otherwise the shortest distance is zero or the parallel-plane formula applies instead.
2. Form the common-perpendicular direction\vec{b_1}\times\vec{b_2} is perpendicular to both direction vectors.
3. Project the joind=\dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} — absolute scalar projection of a point-to-point join onto that common perpendicular.